Let f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3, where | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$. Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$, we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a

Vedclass · Accepted Answer

only if $a_1=0$

Vedclass · Answer

(C) Given $f(x) = a_0 + a_1|x| + a_2|x|^2 + a_3|x|^3$. Since $|x|^2 = x^2$ and $|x|^3 = |x| \cdot x^2$, we can write $f(x) = a_0 + a_1|x| + a_2x^2 + a_3|x|x^2$. For $f(x)$ to be differentiable at $x=0$, the left-hand derivative and right-hand derivative must be equal. Right-hand derivative at $x=0$: $f'(0^+) = \lim_{h 	o 0^+} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{a_0 + a_1h + a_2h^2 + a_3h^3 - a_0}{h} = \lim_{h 	o 0^+} (a_1 + a_2h + a_3h^2) = a_1$. Left-hand derivative at $x=0$: $f'(0^-) = \lim_{h 	o 0^-} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{a_0 + a_1(-h) + a_2h^2 + a_3(-h^3) - a_0}{h} = \lim_{h 	o 0^-} (-a_1 + a_2h - a_3h^2) = -a_1$. For differentiability, $f'(0^+) = f'(0^-) \implies a_1 = -a_1 \implies 2a_1 = 0 \implies a_1 = 0$. Thus, $f(x)$ is differentiable at $x=0$ only if $a_1=0$.

Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$, where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$ if and only if:

Similar Questions

The set of all points,where the derivative of the function $f(x) = \frac{x}{1+|x|}$ exists,is

If $f(x) = \begin{cases} x \left(1 + \frac{1}{2} \sin (\log x^2) \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then find the value of $\lim_{x \rightarrow 0} \frac{f(x) - f(0)}{x}$.

If the function $g(x)=\begin{cases} K \sqrt{x+1} &, 0 \leq x \leq 3 \\ mx+2 &, 3 < x \leq 5 \end{cases}$ is differentiable, then $K+m=$

If $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \ge 0 \end{cases}$ is differentiable at $x = 0$,then $(a, b)$ is

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module